Lesson: Squares and Square Roots

Exercise: 6.1

Question 1

What will be the unit digit of the squares of the following numbers?

(i) 81 (ii) 272 (iii) 799 (iv) 3853 (v) 1234

(vi) 26387 (vii) 52698 (viii) 99880 (ix) 12796 (x) 55555

Answer:

(i) 1 (ii) 4 (iii) 1 (iv) 9 (v) 6 (vi) 9

(vii) 4 (viii) 0 (ix) 6 (x) 5

Solution:

(i) 1 (ii) 4 (iii) 1 (iv) 9 (v) 6

(vi) 9 (vii) 4 (viii) 0 (ix) 6 (x) 5

Question 2

The following numbers are obviously not perfect squares. Give reason.

(i) 1057 (ii) 23453 (iii) 7928 (iv) 222222 (v) 64000

(vi) 89722 (vii) 222000 (viii) 505050

Answer:

(i) Numbers ending in 7 are not square numbers.

(ii) Numbers ending in 3 are not square numbers.

(iii) Numbers ending in 8 are not square numbers.

(iv) Numbers ending in 2 are not square numbers.

(v) Numbers ending in odd number of zeros are not square numbers.

(vi) Numbers ending in 2 are not square numbers.

(vii) Numbers ending in odd number of zeros are not square numbers.

(viii) Numbers ending in odd number of zeros are not square numbers.

Solution:

(i) Numbers ending in 7 are not square numbers.

(ii) Numbers ending in 3 are not square numbers.

(iii) Numbers ending in 8 are not square numbers.

(iv) Numbers ending in 2 are not square numbers.

(v) Numbers ending in odd number of zeros are not square numbers.

(vi) Numbers ending in 2 are not square numbers.

(vii)Numbers ending in odd number of zeros are not square numbers.

(viii) Numbers ending in odd number of zeros are not square numbers.

Question 3

The square of which of the following would be odd numbers?

(i) 431 (ii) 2826 (iii) 7779 (iv) 82004

Answer:

(i) and (iii)

Solution:

(i) and (iii)

Question 4

Observe the following pattern and find the missing digits:

2

11 121

1 01 10201

1 001 1002001

1 00001 1......2.......1

1 0000001 .......................



Answer:

10000200001, 100000020000001

Solution:

2

100001 10000200001

2

10000001 100000020000001

Question 5

Observe the following pattern and find the missing digits:

2

11 121

1 01 10201

1 0101 102030201

1 010101 ....................

...... 10203040504030201



Answer:

1020304030201, 101010101

2

Solution:

2

1010101 1020304030201

2

101010101 10203040504030201

Question 6

Using the given pattern, find the missing numbers:

2 2 2 2

22

1 2 2 3

2 3 6 7

3 4 1 2 1 3

4 5

  



2

2 2 2

2 2 2 2

_2 21

5 _2 30 31

6 7 _ _



  

Answer:

20, 6, 42, 43

Solution:

2 2 2 2

4 5 20 21  

2 2 2 2

5 6 30 31  

2 2 2 2

6 7 42 43  

Question 7

Without adding, find the sum:

(i)

1 3 5 7 9   

(ii)

1 3 5 7 9 11 13 15 17 19        

(iii)

1 3 5 7 9 11 13 15 17 19 21 23          

Answer:

(i) 25 (ii) 100 (iii) 144

Solution:

(i)

2

1 3 5 7 9 5    

[Sum of first five odd natural numbers] = 25

(ii)

1 3 5 7 9 11 13 15 17 19        

= (10)

2

[Sum of first ten odd natural numbers]

= 100

(iii)

1 3 5 7 9 11 13 15 17 19 21 23          

= (12)

2

[Sum of first twelve odd natural numbers]

= 144

Question 8

(i) Express 49 as the sum of 7 odd numbers.

(ii) Express 121 as the sum of 11 odd numbers.

Answer:

(i)

1 3 5 7 9 11 13     

(ii)

1 3 5 7 9 11 13 15 17 19 21         

Solution:

(i)

 

2

49 7 Sum of first 7 odd numbers 1 3 5 7 9 11 13        

 

2

(ii) 121 11 Sum of first 11 odd numbers 1 3 5 7 9 11 13 15 17 19 21            

Question 9

How many numbers lie between squares of the following numbers?

(i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100

Answer:

(i) 24 (ii) 50 (iii) 198

Solution:

(i) Here,

12n 

and

1 13n 



2n non perfect square numbers lie between

 

2

& 1 .nn

Thus,

2 12 24

such numbers lie between 12

2

and 13

2

,

(ii) Similarly,

2 25 50

such numbers lie between 25

2

and 26

2

.

(iii) Similarly,

2 99 198

such numbers lie between 99

2

and 100

2

.

Exercise 6.2

Question 1

Find the square of the following numbers without actual multiplication:

(i) 32 (ii) 35 (iii) 86 (iv) 93 (v) 71 (vi) 46

Answer:

(i) 1024 (ii) 1225 (iii) 7396 (iv) 8649 (v) 5041 (vi) 2116

Solution:

(i)

         

2

32 30 2 30 2 30 2 30 30 2 2 30 2        

22

30 60 60 2 900 120 4 1024       

(ii)

 

2

35 3 4 hundreds 25 12 100 25 1200 25 1225        

(iii)

         

2

86 80 6 80 6 80 6 80 80 6 6 80 6        

22

80 80 6 6 80 6 6400 480 480 36 7396          

(iv)

         

2

93 90 3 90 3 90 3 90 90 3 3 90 3         

2 2

90 90 3 3 90 3 8100 270 270 9 8649          

(v)

         

2

71 70 1 70 1 70 1 70 70 1 1 70 1        

2

70 70 1 1 70 12 4900 70 70 1 5041          

(vi)

         

2

46 40 6 40 6 40 6 40 40 6 6 40 6        

22

40 40 6 6 40 6     

1600 240 240 36 2116    

Question 2

Write a Pythagorean triplet whose one member is:

(i) 6 (ii) 14 (iii) 16 (iv) 18

Answer:

(i) 6,8,10 (ii) 14, 48, 50 (iii) 16, 63, 65 (iv) 18, 80, 82

Solution:

We have the Pythagorean triplet

 

22

2 , – 1, 1 ;m m m 

where

1m 

(natural number).

(i)

2 2

– 1 6 6 1 7Let m m    

Or,

7m 

(Not a natural number)

2

1 6m 

2

6 – 1 5m  

Or,

5m 

(not a natural number)

And

2 6 m 

3m

(A natural number > 1)

Thus,

22

– 1 3 – 1 9 – 1 8m   

and

22

1 3 1 9 1 10m      

Hence, the required triplet is 6, 8, and 10

(ii) Let

2

– 1 14 m 

2

14 1 15m   

Or,

15 m 

(not a natural number)

2

1 14 m 

2

14 – 1 13m  

Or,

13m 

(not a natural number)

And

2 14 m 

7 m

(A natural number> 1)

Thus,

22

– 1 7 – 1 49 – 1 48m   

and

22

1 7 1 49 1 50m      

Hence, the required triplet is 14, 48, and 50

(iii) Let

2

– 1 16 m 

2

16 1 17m   

Or,

17m 

(not a natural number)

or,

2

1 16 m 

2

16 – 1 15 m  

Or,

15m 

(not a natural number)

And

2 16 m 

8m

(A natural number> 1)

Thus,

22

– 1 8 – 1 64 – 1 63m   

and

22

1 8 1 64 1 65m      

Hence, the required triplet is 16, 63, and 65.

(iv) Let

22

– 1 18 18 1 19mm    

Or,

19m 

(not a natural number)

2

1 18 m 

2

18 – 1 17m  

Or,

17m 

(not a natural number) and

2 18 m 

9 m

(A natural number> 1)

Thus,

22

– 1 9 – 1 81 – 1 80m   

and

22

1 9 1, 81 1 82m     

Hence, the required triplet is 18, 80, and 82.

Exercise 6.3

Question 1

What could be the possible ‘one’s’ digits of the square roots of each of the

following numbers?

(i) 9801 (ii) 99856 (iii) 998001 (iv) 657666025

Answer:

(i) 1, 9 (ii) 4, 6 (iii) 1, 9 (iv) 5

Solution:

(i) Unit’s digit of square root of 9801 is 1 or 9.

(ii) Unit’s digit of square root of 99856 is 4 or 6.

(iii) Unit’s digit of square root of 998001 is 1 or 9.

(iv) Unit’s digit of square root of 657666025 is 5.

Question 2

Without doing any calculation, find the numbers which are surely not perfect

squares.

(i) 153 (ii) 257 (iii) 408 (iv) 441

Answer:

(i), (ii), (iii)

Solution:

(i) 153 (ii) 257 (iii) 408 are surely not perfect squares.

Because the numbers that end

with 2, 3, 7 or 8 can never be a perfect square.

Question 3

Find the square roots of 100 and 169 by the method of repeated subtraction.

Answer:

10, 13

Solution:

For 100,

100 – 1 99

99 – 3 96

96 – 5 91

91 – 7 84

84 – 9 75

75 – 11 64

64 – 13 51

51 – 15 36

36 – 17 19

19 – 19 0

We observe that the number 100 got reduced to zero after subtracting 10

consecutive odd numbers starting from 1.

Thus, 100 is a perfect square.

2

100 10

2

100 10 10  

For 169

169 – 1 168

168 – 3 165

165 – 5 160

160 – 7 153

153 – 9 144

144 – 11 133

133 – 13 120

120 – 15 105

105 – 17 88

88 – 19 69

69 – 21 18

48 – 23 25

25 – 25 0

We observe that the number 169 got reduced to zero after subtracting 13

consecutive odd numbers starting from 1.

Thus, 169 is a perfect square.

2

169 13

2

169 13 13  

Question 4

Find the square roots of the following numbers by the Prime Factorization

Method:

(i) 729 (ii) 400 (iii) 1764 (iv) 4096 (v) 7744

(vi) 9604 (vii) 5929 (viii) 9216 (ix) 529 (x) 8100

Answer:

(i) 27 (ii) 20 (iii) 42 (iv) 64 (v) 88 (vi) 98

(vii) 77 (viii) 96 (ix) 23 (x) 90

Solution:

(i) 729

729 3 3 3 3 3 3     

729 3 3 3   

27

(ii) 400

400 2 2 2 2 5 5     

400 2 2 5 20.    

(iii) 1764

1764 2 2 3 3 7 7     

1764 2 3 7 42.    

(iv) 4096

4096 2 2 2 2 2 2 2 2 2 2 2 2           

4096 2 2 2 2 2 2 64.       

(v) 7744

7744 11 11 2 2 2 2 2 2        

7744 11 2 2 2 88.     

(vi) 9604

9604 2 2 7 7 7 7     

9604 2 7 7 98.    

(vii) 5929

5929 7 7 11 11   

5929 7 11 77.   

(viii) 9216

9216 2 2 2 2 2 2 2 2 2 2 3 3           

9216 2 2 2 2 2 3 96.       

(xi) 529

529 23 23 

529 23

(x) 8100

8100 3 3 3 3 2 2 5 5       

8100 3 3 2 5 90.     

Question 5

For each of the following numbers, find the smallest whole number by which it

should be multiplied so as to get a perfect square number.

Also, find the square

root of the square number so obtained.

(i) 252 (ii) 180 (iii) 1008

(iv) 2028 (v) 1458 (vi) 768

Answer:

(i) 7; 42 (ii) 5; 30 (iii) 7, 84 (iv) 3; 78 (v) 2; 54 (vi) 3; 48

Solution:

(i)

252 2 2 3 3 7    

Here, prime 7 does not occur in a pair.



252 is not a perfect square.

If we multiply the given number by 7, then

252 7 2 2 3 3 7 7 1764        

Now, each prime occurs in a pair

252 7 1764 

is a perfect square.

Thus, the required smallest number is 7.

Also,

1764 2 3 7 42.    

(ii) Thus,

180 2 2 3 3 5    

Here, prime 5 does not occur in a pair.



180 is not a perfect square.

If we multiply the given number by 5, then

1 80 5 900 2 2 3 3 5 5        

Now, each prime occurs in a pair.



180 5, . ., 900ie

is a perfect square.

Thus, the required smallest number is 5.

Also,

900 2 3 5 30.   

(iii) 1008

Thus,

1008 2 2 2 2 3 3 7      

Here, prime 7 does not occur in a pair.



1008 is not a perfect square.

If we multiply the given number by 7, then

1008 7 7056 2 2 2 2 3 3 7 7         

Now, each prime occurs in a pair.

1008 7 . ., 7056ie

is a perfect square.

Thus, the required smallest number is 7.

Also,

7056 2 2 3 7 84.    

(iv) 2028

Thus,

2028 2 2 3 13 13    

Here, prime 3 does not occur in a pair.



2028 is not a perfect square.

If we multiply the given number by 3, then

2028 3 6084 2 2 3 3 13 13        

Now, each prime occurs in a pair.

2028 3 . ., 6084ie

is a perfect square.

Thus, the required smallest number is 3.

Also,

6084 2 3 13 78   

(v)

1458 2 3 3 3 3 3 3      

Here, prime 2 does not occur in a pair.



1458 is not a perfect square.

If we multiply the given number by 2, then

1458 2 2916 2 2 3 3 3 3 3 3         

Now, each prime occurs in a pair.

1 458 2, i.e, 2916

is a perfect square.

Thus, the required smallest number is 2.

Also,

2916 2 3 3 3 54.    

(vi)

768 2 2 2 2 2 2 2 2 3        

Here, prime 3 does not occur in a pair.



768 is not a perfect square.

If we multiply the given number by 3, then

768 3 2304 2 2 2 2 2 2 2 2 3 3           

Now, each prime occurs in a pair.

768 3, i.e, 2304

is a perfect square.

Thus, the required smallest number is 3.

Also,

2304 2 2 2 2 3 48.      

Question 6

For each of the following numbers, find the smallest whole number by which it

should be divided so as to get a perfect square.

Find the square root of the square number so obtained.

(i) 252 (ii) 2925 (iii) 396 (iv) 2645

(v) 2800 (vi) 1620

Answer:

(i) 7; 6 (ii) 13; 15 (iii) 11; 6 (vi) 5; 23 (v) 7; 20 (vi) 5; 18

Solution:

(i) 252

252 2 2 3 3 7    

If we divide 252 by 7, then

252 7 36 2 2 3 3     

This is a perfect square.



The required smallest number is 7.

Also,

36 2 3 6  

(ii) 2925

2925 3 3 5 5 13    

If we divide 2925 by 13, then

2925 13 225 3 3 5 5     

This is a perfect square.



The required smallest number is 13.

Also,

225 3 5 15.  

(iii) 396

396 2 2 3 3 11    

If we divide 396 by 11, then

396 11 36 2 2 3 3     

This is a perfect square.



The required smallest number is 11.

Also,

36 2 3 6  

(iv) 2645

2645 5 23 23  

If we divide 2645 by 5, then

2645 5 529 23 23    

which is a perfect square.



The required smallest number is 5.

Also,

529 23

(v) 2800

2800 2 2 2 2 5 5 7      

If we divide 2800 by 7, then

2800 7 400 2 2 2 2 5 5,        

Which is a perfect square.



The required smallest number is 7.

Also,

400 2 2 5 20.    

(vi) 1620

1620 2 2 3 3 3 3 5      

If we divide 1620 by 5, then

1620 5 324 2 2 3 3 3 3        

This is a perfect square.



The required smallest number is 5.

Also,

324 2 3 3 18.    

Question 7

The students of Class VIII of a school donated Rs. 2401 in all, for prime Minister’s

National Relief Fund.

Each student donated as many rupees as the number of students in the class.

Find the number of students in the class.

Answer:

49

Solution:

2401 7 7 7 7   

2401 7 7 49    

Thus, the number of students in

Class VIII is 49.

Question 8

2025 plants are to be planted in a garden in such a way that each row contains as

Many plants as the number of rows.

Find the number of rows and the number of

Plants in each row.

Answer:

45, 45

Solution:

Let the number of plants contained in each row be x.

Then, the number of rows will also be x.



According to the given condition of the question.

2

2025 2025x x x   

 

2

2 2 2 2 2

5 3 3 5 3 3xx     

22

45 45xx

Thus, the number of rows and the number of plants in each row is 45.

Question 9

Find the smallest square number that is divisible by each of the number 4, 9, and

10.

Answer:

900

Solution:

This has to be done in three steps.

Step.1: First find the LCM of the numbers

4, 9 and 10

LCM of 4, 9 and

10 180

Step 2: Multiply the number 180 by 5 to make

it a perfect square.

180 5 2 2 3 3 5 5      

900 2 2 3 3 5 5     

Now, 900 is obviously a perfect square.

Thus, the required square number is 900.

180 2 2 3 3 5    

Question 10

Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.

Answer:

3600

Solution:

LCM of 8, 15 and

20 120.

Prime factorization of

120 2 2 2 3 5    

We see that 2, 3 and 5 are not in pairs.

Therefore, 120 should be multiplied by

2 3 5, . ., 30ie

in order to get a perfect square.

Hence, the required square number

120 30 3600.  

Exercise 6.4

Question 1

Find the square root of each of the following numbers by division method:

(i) 2304 (ii) 4489 (iii) 3481 (iv) 529

(v) 3249 (vi) 1369 (vii) 5776 (viii) 7921

(ix) 576 (x) 1024 (xi) 3136 (xii) 900

Answer:

(i) 48 (ii) 67 (iii) 59 (iv) 23 (v) 57

(vi) 37 (vii) 76 (viii) 89 (ix) 24 (x) 32

(xi) 56 (xii) 30

Solution:

(i)

2304 48

(ii)

4489 67

(iii)

3481 59

(iv)

529 23

(v)

3249 57

(vi)

1369 37

(vii)

5776 76

(viii)

7921 89

(ix)

576 24

(x)

1024 32

(xi)

3136 56

(xii)

900 30

Question 2

Find the number of digits in the square root of the following numbers

(Without any calculation):

(i) 64 (ii) 144 (iii) 4489 (iv) 27225 (v) 390625

Answer:

(i) 1 (ii) 2 (iii) 2 (iv) 3 (v) 3

Solution:

(i) 64: The number of digits in the square root is 1.

(ii) 144: The number of digits in the square root is 2.

(iii) 4489: The number of digits in the square root is 2.

(iv) 27225: The number of digits in the square root is 3.

(v) 390625: The number of digits in the square root is 3.

Question 3

Find the square root of the following decimal numbers:

(i) 2.56 (ii) 7.29 (iii) 51.84 (iv) 42.25 (v) 31.36

Answer:

(i) 1.6 (ii) 2.7 (iii) 7.2 (iv) 6.5 (v) 5.6

Solution:

(i)

2.56 1.6

(ii)

7.29 2.7

(iii)

51.84 7.2

(iv)

42.25 6.5

(v)

31.36 5.6

Question 4

Find the least number which must be subtracted from each of the following numbers

so as to get a perfect square.

Also find the square root of the perfect square

so obtained.

(i) 402 (ii) 1989 (iii) 3250 (iv) 825 (v) 4000

Answer:

(i) 2; 20 (ii) 53; 44 (iii) 1; 57 (iv) 41; 28 (v) 31; 63

Solution:

(i) Here, we get 2 as remainder. It shows that 20

2

is less than 402 by 2.

Thus, the number to be subtracted so as to make it a perfect square is 2.



Required number

402 – 2 400

And

400 20.

(ii) Here, we get remainder 53. It shows that 44

2

is less than1989 by 53.

Thus, the number to be subtracted so as to make it a perfect square is 53.



Required number

1989 – 53 1936 

And

1936 44

\

(iii) Here, we get the remainder 1. It shows that 57

2

is less than 3250 by 1.

Thus, the number to be subtracted so as to make a perfect square is 1.



Required number

3250 – 1 3249

and

3249 57

(iv) Here, we get the remainder 41. It shows that 28

2

is less than825 by 41.

Thus, the number to be subtracted so as to make a perfect square is 41.



Required number

825 – 41 784

And

784 28

(v) Here, we get the remainder 31.

Thus, the smallest number to be

subtracted so as to make a perfect square is 31.



Required number

4000 – 31 3969

And

3969 63

Question 5

Find the least number which must be added to each of the following numbers so as

to get a perfect square.

Also, find the square root of the perfect square so obtained.

(i) 525 (ii) 1750 (iii) 252 (iv) 1825 (v) 6412

Answer:

(i) 4; 23 (ii) 14; 42 (iii) 4; 16 (iv) 24; 43 (v) 149; 81

Solution:

(i) Remainder = 41

2

22 525

Next perfect square number

2

23 529.

Hence, the number to be added

529 – 525 4. 



The required number

525 4 529   

And

529 23

(ii) Remainder = 69

2

41 1750 

Next perfect square number

2

42 1764.

Hence, the number to be added

1760 – 1750 14 



The required number

1750 14 1764  

And

1764 42

(iii) Remainder = 27

2

15 252

Next perfect square number

2

16 256.

Hence, the number to be added

256 – 252 4.



The required number

252 4 256  

And

256 16

(iv) Remainder = 61

2

42 1825

Next perfect square number

2

43 1849.

Hence, the number to be added

1849 – 1825 24.



The required number

1825 24 1849   

And

1849 43

(v) Remainder = 12

2

80 6412 

Next perfect square number

2

81 6561. 

Hence, the number to be added

6561 – 6412 149 



The required number

6412 149 6561   

And

6561 81

Question 6

Find the length of the side of a square whose area is 441 m

2

.

Answer:

21m

Solution:

Let the side of the square be x m.

22

Area m x

i.e.,

2

441x 

Or,

21x 

Thus, the required length of square is 21 m.

Question 7

In a right triangle

ABC, B 90 .

(i)

If AB = 6 cm, BC = 8 cm, find AC.

(ii)

If AC = 13 cm, BC = 5 cm, find AB.

Answer:

   

i 10 cm ii 12 cm

Solution:

 

i Here, AB = 6 cm, BC = 8 cm.

Using Pythagoras’ theorem, we have:

2 2 2

AC = AB + BC

2 2 2

or,AC (36 64) cm 100 cm  

Or,

AC = 10 cm.

(ii) Here,

AC = 13 cm, BC = 5 cm

Using Pythagoras’ theorem, we have:

2 2 2

AC = AB + BC

2 2 2 2 2

Or,AB AC BC (169 25) cm 144 cm    

Or,

AB = 12 cm.

Question 8

A gardener has 1000 plants.

He wants to plant these in such a way that the number of rows and the number of

columns remain the same.

Find the minimum number of more plants he needs for this.

Answer:

24

Solution:

1024 32 32

is a perfect square nearest to 1000



More number of plants he needs

1024 – 1000 24

Question 9

There are 500 children in a school.

For a P.T. drill they have to stand in such a

Manner that the number of rows is equal to the number of columns.

How many children would be left out in this arrangement?

Answer:

16

Solution:

484 22 22

is a perfect square nearest to 500



No. of children left out

500 – 484 16